Problem: The grades on a chemistry midterm at Oak are normally distributed with $\mu = 81$ and $\sigma = 4.0$. Vanessa earned a n $87$ on the exam. Find the z-score for Vanessa's exam grade. Round to two decimal places.
Explanation: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Vanessa's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{87 - {81}}{{4.0}}} $ ${ z \approx 1.50}$ The z-score is $1.50$. In other words, Vanessa's score was $1.50$ standard deviations above the mean.